Principle of Mathematical Induction

Q1.Prove: for n ≥ 1, 5 divides 4ⁿ + 15n – 1. The inductive step uses:

• A4ᵏ⁺¹ + 15(k+1) – 1 = 4(4ᵏ + 15k – 1) – 45k + 14
• B4ᵏ⁺¹ + 15(k+1) – 1 = 4(4ᵏ + 15k – 1) + 15 – 45k
• C4ᵏ⁺¹ + 15k + 14 = 4(4ᵏ + 15k – 1) + 5(–11k+6)
• DCannot be proved by simple induction

Q2.For P(n): 4ⁿ – 1 is divisible by 3, in the inductive step we write 4ᵏ⁺¹ – 1 as:

• A4(4ᵏ – 1) + 3
• B4ᵏ – 1 + 4
• C4ᵏ⁺¹ + 3
• D4ᵏ + 3

Q3.Prove: Σ (2k–1) from k=1 to n equals n². For n = 6, the sum 1+3+5+7+9+11 equals:

• A36 = 6²
• B42
• C30
• D32

Q4.The principle that if P(k) ⟹ P(k+1) and P(1) is true, then P(n) is true for all n ≥ 1, is analogous to:

• AFalling dominoes in a line
• BRolling a die
• CClimbing a staircase randomly
• DDrawing a card from a deck

Q5.For n = 5, the value of n(n+1)(2n+1)/6 is:

• A55
• B45
• C60
• D50